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A router is emitting out IP packets whose total length (data plus header) is $1024$ bytes.Assuming that packets live for $10$ sec, what is the maximum line speed the router can operate at without danger of cycling through the IP datagram identification number space?

1. $53.6$ Mbps
2. $3.5$ Tbps
3. $5.36$ Mbps
4. $35$ Tbps
edited | 229 views
+2

there are 16 bits in the identification number space of an IPV4 datagram

since it is said the lifetime of a packet is 10 sec. So in order to prevent wrap around, at max we can send 216  packets in 10seconds, i.e. in 1 sec, we can send 216/10 packets.

length of a packet is given as 1024B. So in 1 sec, we can send 216 * 1024/10 Bytes. So in 1 sec, we can send 6.4MB. So maximum data rate that is possible is 6.4MBps= (6.4*8)Mbps= 51.2Mbps.

What is the answer for this?

Total length of 1 packet=1024 bytes.

Packet lifetime is 10 seconds , means if a packet with Identification number 1 is produced at time "t" then only at t+10 the same packet with identification number 1 should be produced.

Idenfication number space available is $2^{16}$

Means, the router should be able to exhaust all Identification numbers in 10 seconds only.

1 Identification number is given to a packet of size->1024 Bytes.

So, Router transmission rate must be $\frac{2^{16} \times 1024 \times 8}{10}=53.68Mbps$
edited
+1
Ayush it should be:-

Packet lifetime is 10 seconds, means if a packet with Identification number 1 is produced at time "t" then only at t+10 the same identification number (i.e. 1) should be produced.

We know that total Sequence no. is 2^16 (IPV4)

and packet Size is 1024 bytes

And Time is =10 seconds

Transmission Time  = Length of packet/ Bandwidth

Total Size of packet = 2^16(Total available identification no.) x 1024 (Size of Singles packet )

Bandwidth=  Total Size  of packet / Time

Bandwidth =((2^16)1024*8)/10=53.68Mbps

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I didn't get your logic, how you reached to the answer :/

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